The decay time is given in its proper time, and thus . Solving for v, one finds that choice (D) works.

Alternate Solutions

danielkwalsh2010-09-05 18:25:18

A quick trick: the equation is correct, but for an extremely relativistic particle, as we clearly have here, the decay time in the lab frame can be well approximated as , since . This simplifies the math a lot more in solving for . This gives , which is closest to choice (D).

A quick trick: the equation is correct, but for an extremely relativistic particle, as we clearly have here, the decay time in the lab frame can be well approximated as , since . This simplifies the math a lot more in solving for . This gives , which is closest to choice (D).

To make this faster (avoid the "messy" fraction)
Note to start that this is not a photon thus answer E is out.
now as above,
L=v t
L =v t
some simplification steps=
now plug and chug...=
Note now that we have a
in the first term in the denominator, leaving only
in the denominator,
but
so we simplify<
after canceling we see that
v<
but only by a very small amount thus we have D

hey, I don't understand why t0 is 3 x 10^-8. it seems like I thought t in the frame of the particle was 10^-8 s and that (if you assume the particle to essentially be moving at light speed) then t in the lab frame equals 30/c= 1x 10^-7. any help?